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Polynomial splits over simple extension implies splitting field?

  1. Feb 7, 2016 #1
    This is a question that came about while I attempting to prove that a simple extension was a splitting field via mutual containment. This isn't actually the problem, however, it seems like the argument I'm using shouldn't be exclusive to my problem. Here is my attempt at convincing myself that the subject line is true:

    Let ##p## be a polynomial over a base field ##\mathbb{F}##. Let ##\mathbb{K}=\mathbb{F}(\alpha)## where ##\alpha \not\in\mathbb{F}## and ##\alpha## is a root of p. Now, suppose ##p## splits over ##\mathbb{K}##. Then ##\mathbb{K}## is the splitting field of ##p##. (?)

    Proof. Let ##\mathbb{L}## be the splitting field of ##p## over ##\mathbb{F}##. Then ##\mathbb{L} \subseteq \mathbb{K}## since ##p## splits over ##\mathbb{K}##. Now, ##\alpha \in \mathbb{L}##, but by definition ##\mathbb{K} =\mathbb{F}(\alpha)## is the smallest field containing ##\mathbb{F}## and ##\alpha##. So ##\mathbb{K} \subseteq \mathbb{L}##. So mutual containment is established and ##\mathbb{L} = \mathbb{K}##. //

    Is this argument sound? It seems like if it were, this is something that might have been mentioned in our discussion of splitting fields.
     
  2. jcsd
  3. Feb 7, 2016 #2

    andrewkirk

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    How do you get that ##\alpha\in L##?
     
  4. Feb 8, 2016 #3
    Since ##\alpha## is a root of f and ##\mathbb{L}## is the splitting field of f. ##\mathbb{L}## must contain all roots of f, basically by definition of splitting field.
     
  5. Feb 8, 2016 #4

    andrewkirk

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    I was thinking of the case where ##\mathbb{L}## contains a set of roots, none of which is ##\alpha##, but which is the isomorphic image of a set of roots containing ##\alpha##. I guess that's not a problem if we interpret ##\mathbb{L}## and ##\mathbb{K}## to refer to equivalence classes under isomorphism and the expression ##\mathbb{L}\subseteq\mathbb{K}## to mean for every element of the equivalence class ##\mathbb{L}## there is is an element of ##\mathbb{K}## of which it is a subset.

    If we go that way, your whole argument appears sound to me.
     
  6. Feb 9, 2016 #5
    That is a fair critique I suppose. If we wanted to be super rigorous we could fix an algebraic closure of ##\mathbb{F}## call it ##\mathbb{F}^*## with ##\mathbb{F} \subseteq \mathbb{K} \subseteq \mathbb{F}^*##.

    But either way, while the argument is very short, it seems like a useful lemma that might be discussed at some point. As I find myself making the same argument over and over again. Particularly recently during the discussion of Galois theory.
     
  7. Feb 11, 2016 #6

    mathwonk

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    a filed cannot contain more roots than the degree, so since alpha is contained in K and is a root, and since L is contained in K and contains a full set of roots, then that root alpha must be in L. no fancy algebraic closure choices needed, (although i also first thought of going that route.)
     
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